5
$\begingroup$

According to Theorem 1.7 of Mark-Tosun's paper, the Brieskorn sphere $\Sigma(2,3,6m+1)$ admits two tight contact structure $\xi_{i}\ (i=0,1)$. They are both Stein fillable and they are contactomorphic (but not isotopic).

My question:

Consider the Ozsváth-Szabó contact invariant $c(\xi_{i})\in HF^{+}(-\Sigma(2,3,6m+1))$. We know it is non-zero by Stein fillability, but do we know whether its image under the natural map $$ HF^{+}(-\Sigma(2,3,6m+1))\rightarrow HF^{\text{red}}(-\Sigma(2,3,6m+1)) $$ is non-zero?

Such result can be shown, e.g., if $\xi_{i}$ bounds a non negative definite Stein domain.

$\endgroup$
0

2 Answers 2

3
$\begingroup$

(Note: I might have screwed up orientations, so take everything with a grain of salt.) I will write an argument for the case $m=1$, showing that the reduced contact invariant $c^{\rm red}(\xi)$ is non-zero. I think that the argument(s) can be tweaked to work for $m>1$, as well, but I won't try to write it down properly.

Let me suppress some things from the notation: I'll call $\Sigma = \Sigma(2,3,7)$, and let $\xi$ denote either of $\xi_0$ or $\xi_1$. (The statement is invariant under self-diffeomorphisms of $\Sigma$.)

Mark and Tosun give an explicit surgery presentation of $\xi$ as Legendrian surgery along a right-handed trefoil with Thurston-Bennequin $0$. This gives a Stein cobordism $W: S^3 \leadsto \Sigma$, and the cobordism map $F: = F^+_{\overline W}: HF^+(-\Sigma) \to HF^+(-S^3)$ maps $c(\xi)$ to $c(\xi_{\rm std})$. The key point is showing that $F$ vanishes on the tower in $HF^+(\Sigma)$ (i.e. it vanishes for sufficiently large degrees). This can be seen in two ways.

The first is to show that spin$^c$ structures on $W$ come in conjugate pairs, and all their contributions cancel in pairs on the tower. (Note that they can -and do, a posteriori- move elements outside the tower.)

The second is that $F$ fits into an exact triangle: $$ \dots \to HF^+(-\Sigma_0) \to HF^+(-\Sigma) \to HF^+(-S^3)\to \dots$$ where $\Sigma_0$ is the 0-surgery along the (right-handed) trefoil. It's easy to see (basically by rank count) that the map $HF^+(-\Sigma_0) \to HF^+(-\Sigma)$ must be surjective onto the tower, which must then be killed by $F$.

Either way, if $c^{\rm red}(\xi)=0$, then $F(c^{\rm red}(\xi)) = 0 \neq c(\xi_{\rm std}))$.

$\endgroup$
2
  • $\begingroup$ When we consider the cobordism induced map $F$, don't we need to restrict to the canonical spin-c structure for the symplectic manifold $W$? $\endgroup$
    – user44651
    Oct 19, 2017 at 19:32
  • 1
    $\begingroup$ No, this is Proposition 4.2 in Ozsváth-Szabó's original paper on the contact invariant (which is especially tailored to the purposes here, since the inverse of a Legendrian surgery is a contact +1-surgery). $\endgroup$ Oct 19, 2017 at 20:00
3
$\begingroup$

Brieskorn-Pham calculated the signature of the smoothing of the Milnor fiber of that Brieskorn singularity (in this case the smoothing of the complex singularity $x^2+y^3+z^{6m+1}=0$ or the symplectic branched double cover over the smoothing of the algebraic surface bounding $T(3,7)$ ). In the case of $M(2,3,6m+1)$, this is non-definite. See remark 4.6 of http://www.maths.ed.ac.uk/~aar/papers/nemethi1.pdf for Brieskorn's formula.

In the case of $\Sigma(2,3,6m+1)$, the contact structure has homotopy type $\theta=-2$. As $c_1=0$ for $M(2,3,6m+1)$ (this can be seen by the double cover representation), Gompf's formula for $\theta$, $c_1^2-3\sigma(X)-2\chi(X)=\theta$ gives $$\sigma(M(2,3,6m+1))=-8m$$

$\endgroup$
4
  • $\begingroup$ I at some point had some mathematica code to compute the signature of $M(p,q,r)$ and the homotopy type of the plane field associated to the boundary contact structure which I can scrounge up if any one is interested. $\endgroup$
    – PVAL
    Oct 19, 2017 at 20:32
  • $\begingroup$ When $p=2$, the signature of $M(2,q,r)$ is the Murasugi signature of $T(q,r)$, which is computable in terms of the arithmetics of $q$ and $r$. $\endgroup$ Oct 19, 2017 at 20:42
  • 1
    $\begingroup$ Let me make sure that I understand this right: consider the smooth algebraic curve $C: y^{3}+z^{6m+1}=\epsilon$. After taking double cover of the unit ball $B$ (inside $C^{2}$) branched over $B\cap C$, we get a Stein domain $W$ bounded by $\Sigma(2,3,6m+1)$. The homology of $W$ can be computed (for example, as in L. Kauffman, Open books, branched covers and knot periodicity) and one sees that intersection form of $W$ coincides with the Seifert form of $B\cap C$ (this surface can actually be pushed to a Seifert surface in $S^{3}$), which is not definite. Is this argument right? $\endgroup$
    – user44651
    Oct 19, 2017 at 21:37
  • $\begingroup$ @user44651 That looks right. Brieskorn's original paper computes both $b_2(W)$ and $\sigma(W)$ using a variant of Picard-Lefschetz theory for open manifolds and the Serre spectral sequence. $\endgroup$
    – PVAL
    Oct 19, 2017 at 23:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.